[Java] LeetCode32 Longest Valid Parentheses
2015-04-28 10:02
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Given a string containing just the characters
find the length of the longest valid (well-formed) parentheses substring.
For
which has length = 2.
Another example is
is
题意:给定一串包含括号的字符串,求字长有效的括号串。能顺序匹配的。这一题刚开始想好久~~,用stack来做题,很容易求出字符串中所有有效的括号长度。但是如何求有效子串呢?如果我们能找出无效括号的index,将有效的index减去上一个无效的index,那么就是该字符的有效子串。弄清楚这一点求最大的有效字符串也会变得很容易。
'('and
')',
find the length of the longest valid (well-formed) parentheses substring.
For
"(()", the longest valid parentheses substring is
"()",
which has length = 2.
Another example is
")()())", where the longest valid parentheses substring
is
"()()", which has length = 4.
题意:给定一串包含括号的字符串,求字长有效的括号串。能顺序匹配的。这一题刚开始想好久~~,用stack来做题,很容易求出字符串中所有有效的括号长度。但是如何求有效子串呢?如果我们能找出无效括号的index,将有效的index减去上一个无效的index,那么就是该字符的有效子串。弄清楚这一点求最大的有效字符串也会变得很容易。
public int longestValidParentheses(String s) { if(s==null)return 0; int len=s.length(); int i=0; Stack<Integer> stack=new Stack<Integer>(); char ch; int res=0; while(i<len) { ch=s.charAt(i); if(ch=='(') stack.push(i);//我们变换思路,将括号的index入栈 else { if(!stack.isEmpty()&&s.charAt(stack.peek())=='(')//如果是’)',且与stack顶括号匹配时,弹出 { stack.pop(); res=Math.max(stack.isEmpty()?i+1:i-stack.peek(),res);//为空的话,证明前面没有无效括号,将i+1;不为空的,前面有无效字符,减去无效字符的index }else { stack.push(i); } } i++; } return res; }
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